Problem: You are given N items.
The value of the i-th item (1 \leq i \leq N) is v_i.
Your have to select at least A and at most B of these items.
Under this condition, find the maximum possible arithmetic mean of the values of selected items.
Additionally, find the number of ways to select items so that the mean of the values of selected items is maximized.

Constraints

* 1 \leq N \leq 50
* 1 \leq A,B \leq N
* 1 \leq v_i \leq 10^{15}
* Each v_i is an integer.

Input

The input is given from Standard Input in the following format:


N A B
v_1
v_2
...
v_N


Output

Print two lines.
The first line should contain the maximum possible arithmetic mean of the values of selected items. The output should be considered correct if the absolute or relative error is at most 10^{-6}.
The second line should contain the number of ways to select items so that the mean of the values of selected items is maximized.

Examples

Input

5 2 2
1 2 3 4 5


Output

4.500000
1


Input

4 2 3
10 20 10 10


Output

15.000000
3


Input

5 1 5
1000000000000000 999999999999999 999999999999998 999999999999997 999999999999996


Output

1000000000000000.000000
1

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;

public class Main {
    int n, a, b;
    double[] vs;
    double precision = 10e-6;

    public static void main(String args[]) {
        new Main().run();
    }

    void run() {
        FastReader sc = new FastReader();
        n = sc.nextInt();
        a = sc.nextInt();
        b = sc.nextInt();
        vs = new double[n];
        for (int i = 0; i < n; i++) {
            vs[i] = sc.nextDouble();
        }
        solve();
    }

    void solve() {
        boolean[] maximums = new boolean[n];
        Arrays.sort(vs);
        double sum = 0.0;
        for (int i = n - 1; i >= n - a; i--) {
            sum += vs[i];
        }
        double maxAverage = sum / a;
        double minInAverage = vs[n - a];
        System.out.printf("%6f\n", maxAverage);

        int minCount = 0;
        int index;
        for (index = 0; index < n; index++) {
            if (Math.abs(vs[index] - minInAverage) < precision) {
                minCount++;
            }
            if (vs[index] > minInAverage &&
                    Math.abs(vs[index] - minInAverage) > precision) {
                break;
            }
        }
        int minNum = index - n + a;
        if (minInAverage < maxAverage &&
                Math.abs(minInAverage - maxAverage) > precision) {
            System.out.println(combination(minCount, minNum));
        } else {
            long ans = 0;
            for (int i = minNum; i <= index - n + b; i++) {
                ans += combination(minCount, i);
            }
            System.out.println(ans);
        }
    }

    long combination(int n, int k) {
        long ret = 1;
        for (long i = 1; i <= k; i++) {
            ret *= n + 1 - i;
            ret /= i;
        }
        return ret;
    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements())
            {
                try
                {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException e)
                {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt()
        {
            return Integer.parseInt(next());
        }

        long nextLong()
        {
            return Long.parseLong(next());
        }

        double nextDouble()
        {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try
            {
                str = br.readLine();
            }
            catch (IOException e)
            {
                e.printStackTrace();
            }
            return str;
        }
    }
}